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我整天都在尝试解决“ Learn Python The Hard Way ”一书test_errors()的“练习 48：高级用户输入”中的功能。

assert_equal()，测试中的一个函数要求我按顺序输入元组，但我无法以这种方式对其进行编码。

我的循环总是首先返回名词，最后返回错误元组，我不知道如何打破循环，所以它会重新开始，但使用正确的值继续，或者按照它们应该的顺序对这些元组进行排序所需的任何东西。

这是代码：

class Lexicon(object):

def scan(self, stringo):
    vocabulary = [[('direction', 'north'), ('direction', 'south'), ('direction',     'east'), ('direction', 'west')],
                    [('verb', 'go'), ('verb', 'kill'), ('verb', 'eat')],
                    [('stop', 'the'), ('stop', 'in'), ('stop', 'of')],
                    [('noun', 'bear'), ('noun', 'princess')],    # Remember numbers
                    [('error', 'ASDFADFASDF'), ('error', 'IAS')],
                    [('number', '1234'), ('number','3'), ('number', '91234')]]

    self.stringo = stringo
    got_word = ''
    value = []
    rompe = self.stringo.split() #split rompe en los espacios

    for asigna in vocabulary: 
        for encuentra in asigna:          
            if encuentra[1]  in rompe:
                value.append(encuentra)

    return value   

eLexicon = Lexicon()


from nose.tools import *
from ex48.ex48 import eLexicon 

def test_directions():
    assert_equal(eLexicon.scan("north"), [('direction', 'north')])
    result = eLexicon.scan("north south east")
    assert_equal(result, [('direction', 'north'),
                  ('direction', 'south'),
              ('direction', 'east')])

def test_verbs():
    assert_equal(eLexicon.scan("go"), [('verb', 'go')])
    result = eLexicon.scan("go kill eat")
    assert_equal(result, [('verb', 'go'),
                  ('verb', 'kill'),
                  ('verb', 'eat')])

def test_stops():
    assert_equal(eLexicon.scan("the"), [('stop', 'the')])
    result = eLexicon.scan("the in of")
    assert_equal(result, [('stop', 'the'),
                  ('stop', 'in'),
                  ('stop', 'of')])

def test_nouns():
    assert_equal(eLexicon.scan("bear"), [('noun', 'bear')])
    result = eLexicon.scan("bear princess")
    assert_equal(result, [('noun', 'bear'),
                  ('noun', 'princess')])

#def test_numbers():
#   assert_equal(lexicon.scan("1234"), [('number', 1234)])
#   result = lexicon.scan("3 91234")
#   assert_equal(result, [('number', 3),
#                 ('number', 91234)])

def test_errors():
    assert_equal(eLexicon.scan("ASDFADFASDF"), [('error', 'ASDFADFASDF')])
    result = eLexicon.scan("bear IAS princess")
    assert_equal(result, [('noun', 'bear'),
                  ('error', 'IAS'),
                  ('noun', 'princess')])

======================================================================
FAIL: tests.ex48_tests.test_errors
----------------------------------------------------------------------
Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/nose/case.py", line 197, in runTest
    self.test(*self.arg)
  File "/home/totoro/Desktop/Python/projects/ex48/tests/ex48_tests.py", line 43, in         test_errors
    ('noun', 'princess')])
AssertionError: Lists differ: [('noun', 'bear'), ('noun', 'p... != [('noun', 'bear'),     ('error', '...

First differing element 1:
('noun', 'princess')
('error', 'IAS')

- [('noun', 'bear'), ('noun', 'princess'), ('error', 'IAS')]
+ [('noun', 'bear'), ('error', 'IAS'), ('noun', 'princess')]

----------------------------------------------------------------------
Ran 5 tests in 0.006s

非常感谢您抽出宝贵的时间。

4

4 回答 4

1

这对我来说很好，它是更短的代码。前段时间有这本书，幸好我还有文件...

def check(word):
    lexicon = {
        'direction': ['north', 'south', 'east', 'west'],
        'verb': ['go', 'kill', 'eat'],
        'stop': ['the', 'in', 'of'],
        'noun': ['bear', 'princess'],
        'error': ['ASDFADFASDF', 'IAS']
    }
word = str(word)
for key in lexicon:
    if word in lexicon[key]:
        return (key, word)
    elif word.isdigit():
        return ('number', int(word))

def scan(words):
    words = words.split()
    to_return = []
    for i in words:
        to_return.append(check(i))
    return to_return

这出现了：

......
----------------------------------------------------------------------
Ran 6 tests in 0.008s

OK

告诉我这段代码是否有任何错误。只需在下面评论：D。

于 2015-03-18T12:32:00.060 回答
0

测试中的单词与出来的顺序相同。因此，您需要重新排序for-loops 以首先迭代输入：

    value = []
    for rompe in stringo.split():
        for asigna in vocabulary:
            for encuentra in asigna:
                if encuentra[1] == rompe:
                    value.append(encuentra)

这将以encuentra正确的顺序返回 s。

注意 1：您不应该对数字或错误进行硬编码。

注意 2：您可以通过使用一两个字典来大大降低此算法的复杂性。

例子：

vocabulary = {
    'direction': 'north east south west up down left right back'.split(),
    'noun': 'bear princess door cabinet'.split(),
    'stop': 'the in of from at it'.split(),
    'verb': 'go kill eat stop'.split(),
}

'''
This creates a lookup using a dictionary-comprehension:
{'at': 'stop',
# [...]
 'up': 'direction',
 'west': 'direction'}
'''
classifications = {i: k for k, v in vocabulary.iteritems() for i in v}


def classify(word):
    try:
        return 'number', int(word)
    except ValueError:
        return classifications.get(word, 'error'), word


def scan(words):
    return [classify(word) for word in words.split()]
于 2013-07-06T07:44:00.067 回答
0 
    for word in self.stringo.split(): 
        for pair in vocabulary:             
            if pair[0][1] == word:
                value.append(pair[0])
            elif pair[1][1] == word:
                value.append(pair[1])
            elif pair[2][1] == word:
                value.append(pair[2])
            elif pair[3][1] == word:
                value.append(pair[3])
于 2013-07-06T14:08:36.057 回答
0

我刚刚完成了这个练习，希望这能给你们一些新的想法。这是我的解决方案：

#Set up datastructure
direction = ["north", "east", "south", "west", "up", "right", "down", "left", "back"]
verb = ["go", "stop", "kill", "eat"]
stop = ["the", "in", "of", "from", "at", "it"]
noun = ["door", "bear", "princess", "cabinet"]
vocabulary = [(direction, 'direction'), (verb, 'verb'), (stop, 'stop'), (noun, 'noun')]

def scan(sentence):
    #searches the words in the datastructure, if not found checks if it is an integer, if not returns error.
    results = []
    words = sentence.split()

    for word in words:
        found = False
        for category in vocabulary:
            if word.lower() in category[0]:
                results.append((category[1], word))
                found = True
            else:
                pass
        if found is False and isInt_str(word) is True:
            results.append(('number', int(word)))
        elif found is False and isInt_str(word) is False:
            results.append(('error', word))
        elif found is True:
            pass
        else:
            print("I'm terribly sorry, but something you entered is neither a word nor a number.")
    return results

def isInt_str(string):
    #returns True or False if string equals an integer. (i.e. 2 = True, 2*-2 = True 2**0,5 = False)
    string = str(string).strip()
    return string=='0' or (string if string.find('..') > -1 else string.lstrip('-+').rstrip('0').rstrip('.')).isdigit()
于 2018-03-28T10:33:40.700 回答